The current density in a cylindrical wire of radius r=4.0 mm… - JEE Main 2022 Physics PYQ

JEE Main 2022 — Physics Electromagnetism

2022integermedium

The current density in a cylindrical wire of radius $r=4.0\mathrm{mm}$ is $1.0\times {10}^{6}{A}^{2}{m}^{2}$ . The current through the outer portion of the wire between radial distances $\frac{r}{2}$ and $r$ is $x\pi A$ ; where $x$ is

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Held on 27 Jun 2022 · Verified 6 Jul 2026.

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Solution

Current $=J\times$ Area

$=J\times \pi {[{R}^{2}-(\frac{R}{2})]}^{2} =J\pi \frac{3{R}^{2}}{4} ={10}^{6}\times \pi \times 3\times {(\frac{4\times {10}^{-3}}{4})}^{2}$

$\Rightarrow \pi \times 3\times \frac{16}{4}=12\pi A$

Hence, the value of $x=12$ .

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